Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $y \neq 0$. $x = \dfrac{y^2 - 13y + 30}{-5y - 10} \times \dfrac{2y + 4}{2y - 20} $
Answer: First factor the quadratic. $x = \dfrac{(y - 10)(y - 3)}{-5y - 10} \times \dfrac{2y + 4}{2y - 20} $ Then factor out any other terms. $x = \dfrac{(y - 10)(y - 3)}{-5(y + 2)} \times \dfrac{2(y + 2)}{2(y - 10)} $ Then multiply the two numerators and multiply the two denominators. $x = \dfrac{ (y - 10)(y - 3) \times 2(y + 2) } { -5(y + 2) \times 2(y - 10) } $ $x = \dfrac{ 2(y - 10)(y - 3)(y + 2)}{ -10(y + 2)(y - 10)} $ Notice that $(y + 2)$ and $(y - 10)$ appear in both the numerator and denominator so we can cancel them. $x = \dfrac{ 2\cancel{(y - 10)}(y - 3)(y + 2)}{ -10(y + 2)\cancel{(y - 10)}} $ We are dividing by $y - 10$ , so $y - 10 \neq 0$ Therefore, $y \neq 10$ $x = \dfrac{ 2\cancel{(y - 10)}(y - 3)\cancel{(y + 2)}}{ -10\cancel{(y + 2)}\cancel{(y - 10)}} $ We are dividing by $y + 2$ , so $y + 2 \neq 0$ Therefore, $y \neq -2$ $x = \dfrac{2(y - 3)}{-10} $ $x = \dfrac{-(y - 3)}{5} ; \space y \neq 10 ; \space y \neq -2 $